if sinA+sinB=x and cosA-cosB=y then find tan[(A-B)/2]
Two of the lesser known identities are:
sinA + sinB = 2sin[(A+B)/2]cos[(A-B)/2] and
cosA - cosB = -2sin[A+B)/2]sin[A-B)/2]
so
x/y = ( 2sin[(A+B)/2]cos[(A-B)/2] ) / (-2sin[A+B)/2]sin[A-B)/2] )
x/y = - cos[(A-B)/2] / sin[A-B)/2]
x/y = - cot[(A-B)/2]
tan[(A-B)/2] = -y/x
use your sum-to-product formulas.
sinA + sinB = 2 sin (A+B)/2 cos (A-B)/2
now use the formula for sum of cosines, and recall that
tanθ = sinθ / cosθ
To find the value of tan[(A-B)/2], we first need to simplify the given expressions sinA + sinB = x and cosA - cosB = y.
Let's start by looking at the trigonometric identities:
1. sin(A + B) = sinA*cosB + cosA*sinB
2. cos(A + B) = cosA*cosB - sinA*sinB
3. sin(A - B) = sinA*cosB - cosA*sinB
4. cos(A - B) = cosA*cosB + sinA*sinB
Using equations (1) and (3), we can rewrite the given expressions:
sinA + sinB = x
=> 2*sin((A + B)/2)*cos((A - B)/2) = x
cosA - cosB = y
=> -2*sin((A + B)/2)*sin((A - B)/2) = y
Now, let's divide these two equations:
(x) / (y) = [2*sin((A + B)/2)*cos((A - B)/2)] / [-2*sin((A + B)/2)*sin((A - B)/2)]
The 2 and -2 terms will cancel each other out. We'll also simplify the sine and cosine terms:
(x) / (y) = [-cos((A - B)/2)] / [sin((A - B)/2)]
Now, let's rewrite this in terms of tangent:
(x) / (y) = -cot((A - B)/2)
Multiplying both sides by -1, we get:
(-x) / (y) = cot((A - B)/2)
Finally, taking the reciprocal of both sides, we find:
y / (-x) = tan((A - B)/2)
Therefore, tan[(A - B)/2] = y / (-x).
To find the value of tan[(A-B)/2], we need to use some trigonometric identities and solve for the values of A and B.
Let's start by rearranging the given equations:
sinA + sinB = x ...(1)
cosA - cosB = y ...(2)
To find the value of tan[(A-B)/2], we need to determine the values of sin[(A-B)/2] and cos[(A-B)/2].
Let's use the following identities:
sin[(A-B)/2] = ± √[(1 - cos(A-B))/2] ...(3)
cos[(A-B)/2] = ± √[(1 + cos(A-B))/2] ...(4)
Now, let's eliminate A and B from equations (1) and (2) by squaring them:
(sinA + sinB)^2 = x^2 ...(5)
(cosA - cosB)^2 = y^2 ...(6)
Expanding the equations (5) and (6), we get:
(sin^2A + sin^2B + 2sinA*sinB) = x^2 ...(7)
(cos^2A + cos^2B - 2cosA*cosB) = y^2 ...(8)
Using the pythagorean identity, sin^2A + cos^2A = 1 and sin^2B + cos^2B = 1, we can rearrange equations (7) and (8) as follows:
1 + 2sinA*sinB = x^2 ...(9)
1 - 2cosA*cosB = y^2 ...(10)
Now, let's combine equations (9) and (10) to get an equation in terms of sin(A-B):
1 + 2sinA*sinB = 1 - 2cosA*cosB
Rearranging this equation, we get:
sinA*sinB + cosA*cosB = 0
Applying the trigonometric identity, sinA*sinB = 1/2[cos(A-B) - cos(A+B)], we can simplify the equation further:
1/2[cos(A-B) - cos(A+B)] + cosA*cosB = 0
Expanding and rearranging the equation:
cosA*cosB + 1/2[cos(A-B)] - 1/2[cos(A+B)] = 0
Now, we can substitute cos(A-B) = 2cos^2[(A+B)/2] - 1 and cos(A+B) = 2sin^2[(A+B)/2] - 1:
cosA*cosB + 1/2[2cos^2[(A+B)/2] - 1] - 1/2[2sin^2[(A+B)/2] - 1] = 0
Simplifying further, we get:
cosA*cosB + cos^2[(A+B)/2] - sin^2[(A+B)/2] = 0
Now, substituting sin^2[(A+B)/2] with 1 - cos^2[(A+B)/2], we obtain:
cosA*cosB + cos^2[(A+B)/2] - (1 - cos^2[(A+B)/2]) = 0
Simplifying, we get:
cosA*cosB + 2cos^2[(A+B)/2] - 1 = 0
The equation obtained is a quadratic equation in terms of cos[(A+B)/2]. Solving this quadratic equation will give us the value of cos[(A+B)/2]. Once we have cos[(A+B)/2], we can substitute it back into equation (3) to find sin[(A-B)/2]. Finally, using the formula tan[(A-B)/2] = sin[(A-B)/2] / cos[(A-B)/2], we can determine the value of tan[(A-B)/2].
Please note that solving the quadratic equation involves mathematical calculations that may require additional steps not shown here.
So, the process to find the value of tan[(A-B)/2] involves eliminating A and B from the given equations, expressing the equation in terms of trigonometric identities, solving a quadratic equation for cos[(A+B)/2], substituting the value back into equation (3) to find sin[(A-B)/2], and then calculating tan[(A-B)/2] using the formula tan[(A-B)/2] = sin[(A-B)/2] / cos[(A-B)/2].